Problem: A board game spinner is divided into three regions labeled $A$, $B$ and $C$.  The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$.  What is the probability of the arrow stopping on region $C$?  Express your answer as a common fraction.
Solution: Since the sum of the three probabilities is 1, the probability of stopping on region $C$ is $1 - \frac{1}{3} -
\frac{1}{2} = \frac{6}{6} - \frac{2}{6} - \frac{3}{6} = \boxed{\frac{1}{6}}$.